Curvature. 1. the Curvature Tensor. Let (M,G) Be a Smooth Manifold With

Curvature.

1. The Curvature Tensor. Let (M, g) be a smooth manifold with a smooth Riemannian metric g. Let be the Riemannian connection on M corresponding to g. The curvature tensor∇R is defined as follows. Let X,Y,Z be smooth vector fields on M. Then the curvature tensor R is defined as:

( ) R(X,Y )Z := X Y Z Y X Z Z. † ∇ ∇ − ∇ ∇ − ∇[X,Y ] Thus, given three vector fields X,Y,Z, the curvature tensor outputs another vector field R(X,Y )Z. Facts: (1) Let (U, φ = (x1, . . . , xn)) be a chart on M and let X,Y,Z be vector fields that in this chart have the form ∂ ∂ ∂ X = ai(x) ,Y = bj(x) ,Z = ck(x) . X ∂xi X ∂xj X ∂xk i j k Then ∂ ∂ ∂ R(X,Y )Z = aibjckR( , ) . X ∂xi ∂xj ∂xk i,j,k

Thus (R(X,Y )Z) depends ONLY on X p,Y p,Z p. p | | |

(2) The above property allows us to convert R into an actual 3-covariant 1- contravariant tensor R as follows. Let p M, v1, v2,e v3 Mp and u M ∗. Choose smooth vector fields X,Y,Z ∈ ∈ ∈ p such that X(p) = v1, Y (p) = v2 and Z(p) = v3. Recall that (R(X,Y )Z)(p) depends ONLY on v1, v2, v3 and not on the specific choice of X,Y,Z. Put

R p(u, v1, v2, v3) := u R(Z,Y )X . | 
p In view of (1) this formulae does define a multilinear map

R p : M ∗ Mp Mp Mp R, | p × × × → e so that R is indeed a tensor field on M. (3) Symmetriese of the curvature tensor: (a) R(X,Y )Z = R(Y,X)Z; (b) R(X,Y )Z + R−(Y,Z)X + R(Z,X)Y = 0; (c) g(R(X,Y )Z,V ) = g(R(X,Y )V,Z); (d) g(R(X,Y )Z,V ) = −g(R(Z,V )X,Y ). ∂ ∂ ∂ (4) As, noted above, computing R reduces to computing R( ∂xi , ∂xj ) ∂xk . Define s Rijk as: ∂ ∂ ∂ ∂ R( , ) = Rs . ∂xj ∂xk ∂xi X ijk ∂xs s Note that Rs = dxs R( ∂ , ∂ ) ∂ ijk ∂xj ∂xk ∂xi
Thus Rs = R dxs, ∂ , ∂ , ∂ , which agrees with the standard conventions ijk ∂xi ∂xj ∂xk
for denoting tensore coefficients. 1 2

(5) One can express the coefficients of the curvature tensor directly in terms of the Christoffel symbols:

∂Γs ∂Γs Rs = ij ik + Γr Γs Γr Γs . ijk ∂xk ∂xj X ij rk ik rj − r −
However, in practice it is almost always better to compute R(X,Y )Z via its definition ( ). † 2. Sectional curvature. Let (M n, g) be as in Section 1 (with n 2). For vector fields X,Y on M define ≥ κ(X,Y ) = g (R(X,Y )Y,X) and 2 κ1(X,Y ) = g(X,X)g(Y,Y ) g(X,Y ) . − Let p M and let σ Mp be a two-dimensional linear subspace.Let v1, v2 σ ∈ ≤ ∈ be a basis of σ. Choose X and Y such that X(p) = v1 and Y (p) = v2. Then the sectional curvature Kσ of M at p along σ is defined as:

κ(X,Y ) Kσ := . κ1(X,Y ) Note: On the face of it, the above definition of Kσ depends on the choice of a basis v1, v2 of σ. However, one can show that for any α, β, γ, δ R if X1 = αX +βY ∈ and Y1 = γX + δY then 2 κ(X1,Y1) = (αδ βγ) κ(X,Y ) − and 2 κ(X1,Y1) = (αδ βγ) κ(X1,Y1) − Thus shows that Kσ indeed depends only on σ and not on the choice of a particular basis in σ. 3. Curvature of a space curve. Let c(t) = (x(t), y(t), z(t)) be a smooth regular curve in R3. Put v(t) = c˙(t) , c˙(t) || || the speed of c at time t and put T (t) := v(t) , the unit tangent to c at time t. Then the curvature k(t) of c at time t is defined as: T˙ (t) k (t) = k(t) := || ||. c v(t) Note that by definition k(t) 0 for every t. Also, if c(s) is a unit speed curve, ≥ then T (s) = c0(s) and hence k(s) = T 0(s) = c00(s) . ||1 || || || One can also show that k(t) = r(t) where r(t) is the radius of the osculating circle (which is a circle providing the “best” second order approximation of the curve c tangent to c at the point c(t)) 4. Gaussian curvature. Let M 2 R3 be an oriented surface in R3 with an outward unit normal n. Let ⊆ p M and let v Mp be a nonzero vector. The intersection of the plane through p ∈parallel to v and∈ n(p) is a curve. We can parameterize this curve c(t) so that c(0) = p andc ˙(0) = v Mp. One can define a signed curvature kp(v) := kc(0) where  = 1 if the endpoint∈ of n(p) is on the same side of M as the center of − 3 the osculating circle to c at p and  = 1 otherwise. In other words,  = 1 exactly when n “turns” in the same direction as c(t). The maximum and the minimum values of kp(v) (when p is fixed and v varies over Mp 0 ) are called the principal − { } curvatures of M at p and are denoted k1 and k2. The Gaussian curvature Kp of M at p is Kp = k1k2. It is not hard to see that changing the orientation on M, that is replacing n by n preserves Kp. Moreover, we only need to orient the surface M “locally”, − near the point p in order for Kp to be defined as above. Thus Gaussian curvature can be defined for non-orientable surfaces as well, by choosing “local” outward unit normals. 5. The second fundamental form and Gaussian curvature. Let M 2 R3 be as in Section 4. Let Ω R2 be an open set and f :Ω M be the inverse⊆ of some chart map for M. Thus⊆ f is a parameterization of an→ open subset U = f(Ω) of M. We use (x1, x2) to denote the coordinates in R2. Assume that f is orientation-preserving, so that n is a positive scalar multiple of f 1 f 2 . x × x The second fundamental form is a 2 2 matrix with entries hij, i, j = 1, 2, where × ∂2f ∂n ∂f hij = n, = , h ∂xj∂xi i −h∂xj ∂xi i Note that hij = hji so that the second fundamental form is a symmetric matrix. The traditional notation for its coefficients is L = h11, M = h12 = h21 and N = h22. Recall that gij = f i , f j . h x x i (1) For v = αf 1 + βfx Mp (where α, β R) we have x 2 ∈ ∈ Lα2 + 2Mαβ + Nβ2 Lα2 + 2Mαβ + Nβ2 kp(v) = 2 = 2 2 . v α g11 + 2αβg12 + β g22 || || (2) The above formula can be used (with some extra work) to derive an explicit formula for the Gaussian curvature: Theorem A. For p = f(x1, x2) we have det h K = ij . p det g (x1,x2) ij

6. The shape operator. In the notations of the previous section let p M ∈ and v Mp. Put ∈ S(v) := Dvn. − It is easy to see that S is a linear operator. Note that 1 = n, n and hence h i 0 = Dv1 = Dv n, n = Dvn, n + n,Dvn = 2 Dvn, n . h i h i h i h i Hence Dvn is perpendicular to n, so that S(v) Mp. Thus S : Mp Mp is a linear transformation called the shape operator (or∈ the Weingarten map)→ . Hence the determinant det S of S is well-defined: choose a basis of Mp, write the matrix of S with respect to that basis and take the determinant of that matrix. Choosing another basis of Mp conjugates this matrix (by the transition matrix from one basis to another) but the determinant remains the same. The shape operator can be used to compute the Gaussian curvature, namely: 4

Kp = det S.

7. Sectional curvature and Gaussian curvature. Let M 2 R3 be a surface as in the preceding three sections. Let p M. Note ⊂ ∈ that Mp is two-dimensional, so it has only one two-dimensional subspace, namely σ = Mp itself. It turns out that in this case the Gaussian curvature and the sectional curvature defined in Section 2 via the curvature tensor coincide. Theorem B. We have:

Kp = Kσ.

This result shows that the notion of the Gaussian curvature is intrinsic for a surface M with a Riemannian metric g, since the curvature tensor R and the sec- tional curvature are defined purely in intrinsic terms, rather than using a particular embedding. About the proof of Theorem B. Let p M and f :Ω U M, (where p U and p = f(0)) be a parame- ∈ → ⊆ ∈ ∂f ∂f terization of an open subset of M, as above. We may assume that ∂x1 0, ∂x2 0 is 3 an orthonormal basis of Mp R (we can always achieve this by pre-composing f ≤ 2 with an appropriately chosen linear transformation of R ). Thus gij 0 = δij and | det(gij) 0 = 1. Hence by Theorem A | 2 ( ) Kp = det(hij) 0 = h11h22 h 0 ∗ | − 12
| in this case. Recall an explicit formula for the Riemannian connection on M in this case:

X Y = DX Y DX Y, n n. ∇ − h i We want to compute ∂f ∂f ∂f ∂f ∂f ∂f k( , ) = R( , ) , ∂x1 ∂x2 h ∂x1 ∂x2 ∂x2 ∂x1 i ∂f in this case. Since n is orthogonal to ∂x1 , we can disregard the coefficient at n in ∂f ∂f ∂f ∂f ∂f ∂ ∂ R( ∂x1 , ∂x2 ) ∂x2 . Recall also that [ ∂x1 , ∂x2 ] = [ ∂x1 , ∂x2 ] = 0. We have ∂f ∂2f ∂2f ∂2f ∂f 2 = 2 2 2 2 , n n = 2 2 h22n. ∇ ∂x2 ∂x (∂x ) − h(∂x ) i (∂x ) − Hence ∂f ∂2f ∂f ∂f 2 = ∂f  2 2 h22n = ∇ ∂x1 ∇ ∂x2 ∂x ∇ ∂x1 (∂x ) − ∂3f ∂n h22 + an ∂x1(∂x2)2 − ∂x1 where the coefficient a is unimportant. Similarly, we have: ∂f ∂2f ∂2f ∂2f ∂f 2 = 1 2 1 2 , n n = 1 2 h12n. ∇ ∂x1 ∂x ∂x ∂x − h∂x ∂x i ∂x ∂x − 5

Hence ∂f ∂2f ∂f ∂f 2 = ∂f  1 2 h12n = ∇ ∂x2 ∇ ∂x1 ∂x ∇ ∂x2 ∂x ∂x − ∂3f ∂n h12 + bn ∂x1(∂x2)2 − ∂x2 where the coefficient b is unimportant. Hence ∂f ∂f ∂f ∂n ∂n R( , ) = h12 h22 + (a b)n. ∂x1 ∂x2 ∂x2 ∂x2 − ∂x1 − Then ∂f ∂f ∂f ∂f ∂f ∂f k( , ) = R( , ) , = ∂x1 ∂x2 h ∂x1 ∂x2 ∂x2 ∂x1 i ∂n ∂f ∂n ∂f 2 h12 , h22 , = h + h22h11 = det(hij). h∂x2 ∂x1 i − h∂x1 ∂x1 i − 12 3 ∂f ∂f Let σ = Mp R , so that σ has basis 1 , 2 . Recall that by assumption ≤ ∂x 0 ∂x 0 this is an orthonormal basis of M , so that k ( ∂f , ∂f ) = 1. Hence p 1 ∂x1 0 ∂x2 0 ∂f ∂f k( ∂x1 , ∂x2 ) det(hij) 0 Kσ = =
= Kp k ( ∂f , ∂f ) 0 1 1 ∂x1 0 ∂x2 where the last equality holds in view of ( ). ∗